The precaution to be taken while measuring the temperature of a liquid in a beaker is applying proper heat balance and taking all the required precautions.
- A beaker with an open top contains a sample of liquid. It exposes this sample to light.
- That liquid absorbs the light energy, turning it into heat energy. As a result, the liquid becomes warmer and evaporation is accelerated. As a result, there is less liquid in the beaker.
- Since it is well known that the surface temperature of a liquid, along with air movement above the liquid surface, is one of the dominant factors affecting evaporation, I want to measure the evaporation rate as a function of surface temperature.
- This can be done by applying a heat balance.
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Methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75
Answer:
SEE BELOW
Explanation:
Glucose (C6H12O6) is an organic compound composed of the elements carbon, hydrogen and oxygen.
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
